Rational InequalityDate: 10/09/2001 at 19:57:42 From: alan Subject: Algebra, rational inequality I have a question about how to solve a rational inequality and give an answer in interval notation. The example looks like this. - ____5____ greater than or equal to _5_ 3h + 2 h I do not even know where to start. Please help! Date: 10/10/2001 at 18:12:13 From: Doctor Greenie Subject: Re: Algebra, rational inequality Hi, Alan - Let's find the solution first by looking at the graphs of the two rational expressions separately and analyzing them. Later we can look at a purely algebraic way to reach the same solution. I. Solution by analyzing graphs (1) -5/(3h+2) At h = -2/3, the denominator is zero and the value is undefined. For values of h larger than -2/3, the denominator is positive, so the function value is negative; for values of h less than -2/3, the denominator is negative, so the function value is positive. For large negative values of h, the function value is a small positive number; for large positive values of h, the function value is a small negative number. In summary, the value for this function is - small positive for large negative values of h; - increasingly large positive as you move towards h=-2/3, becoming very large positive for values of h slightly less than -2/3; - undefined for h=-2/3; - very large negative for values of h slightly larger than -2/3; - increasing (less negative) as you move towards values of h approaching infinity; small negative for large positive values of h (2) 5/h At h = 0, the denominator is zero and the value is undefined. For values of h larger than 0, the denominator is positive, so the function value is positive; for values of h less than 0, the denominator is negative, so the function value is negative. For large negative values of h, the function value is a small negative number; for large positive values of h, the function value is a small positive number. In summary, the value for this function is - small negative for large negative values of h; - increasingly large negative as you move towards h=0, becoming very large negative for values of h slightly less than 0; - undefined for h=0; - very large positive for values of h slightly larger than 0; - decreasing (less positive) as you move towards values of h approaching infinity; small positive for large positive values of h With the given inequality, we want to find the values of h for which the value of function (1) is greater than or equal to the value of function (2). (a) Function (1) is positive for all values of h less than -2/3; function (2) is negative for all negative values of h. Therefore, the inequality is satisfied for all values of h less than -2/3. (b) Function (1) is negative for all values of h greater than -2/3; function (2) is positive for all positive values of h. Therefore, the inequality is not satisfied for any values of h greater than 0. (c) In the interval between h-values of -2/3 and 0, both functions are negative; the function (1) value is rapidly increasing from negative infinity towards 0; the function (2) value is decreasing rapidly toward negative infinity. Somewhere in the interval the two function values will be the same; the inequality will be satisfied in the interval between that value and 0. To find the value where the two function values are equal, we solve the equation associated with the given inequality: -5 5 ---- = - 3h+2 h Multiplying both sides of the equation by the two denominators to clear fractions, we have -5h = 5(3h+2) -5h = 15h+10 -20h = 10 h = -1/2 We have determined, by analyzing the graphs of the two functions, that the inequality is satisfied for values of h which are either in (-infinity,-2/3) or in [-1/2,0). The two functions have equal values at h = -1/2, so equality holds there and so h=-1/2 is included in the solution set of the "greater than or equal to" inequality, as indicated by the square bracket. One or the other of the functions is undefined at h=0 and h = -2/3, so those points are not included in the solution set of the inequality, as indicated by the curved brackets. II. Algebraic solution Now let's look at a couple of different ways to arrive at the same solution set algebraically, without analyzing the graphs of the functions. Probably the easiest way to find the solution set to the inequality -5 5 ---- >= - 3h+2 h is to rewrite the inequality as -5 5 ---- - - >= 0 3h+2 h Then we can combine the fractions on the left: -5(h)-5(3h+2) -5h-15h-10 -10(2h+1) ------------- = ---------- = --------- h(3h+2) h(3h+2) h(3h+2) This rational function is undefined (denominator is 0) at h=0 and h=-2/3; it has the value 0 (numerator is 0) when h=-1/2. These three critical points divide the set of possible h-values into four intervals; we need to examine the function in each of those intervals to see whether the function value is positive or negative, as determined by the signs of the two factors in the numerator and the two factors in the denominator. The original inequality is satisfied when this function value is positive. We have (-)(-) (-infinity,-2/3): ------ = (+) YES (-)(-) (-)(-) (-2/3,-1/2): ------ = (-) NO (-)(+) (-)(+) (-1/2,0): ------ = (+) YES (-)(-) (-)(+) (0, infinity): ------ = (-) NO (+)(+) This analysis shows the strict inequality is satisfied on the intervals (-infinity, -2/3) and (-1/2,0); and we know the equality is satisfied at h=-1/2. So, as before, our solution set is (-infinity, -2/3) and [-1/2,0). A less common algebraic approach to the solution would be to multiply the original inequality on both sides by the two denominators. When you do this, you need to remember that if you multiply both sides of an inequality by a negative number, the direction of the inequality is reversed. So again we start with the original inequality -5 5 ---- >= - 3h+2 h and we multiply both sides by (h)(3h+2). The factor (h) is negative for h-values less than 0 and positive for h-values greater than 0; the factor (3h+2) is negative for h-values less than -2/3 and positive for h-values greater than -2/3. The product (h)(3h+2) is therefore (a) positive (negative * negative) for h-values less than -2/3; or (b) negative (negative * positive) for h-values between -2/3 and 0; or (c) positive (positive * positive) for h-values greater than 0 After multiplying the inequality by (h)(3h+2) and reversing the direction of the inequality when needed, we have (a) -5h >= 5(3h+2) and h < -2/3 -5h >= 15h+10 and h < -2/3 -20h >= 10 and h < -2/3 h <= -1/2 and h < -2/3 h < -2/3 or (b) -5h <= 5(3h+2) and -2/3 < h < 0 -5h <= 15h+10 and -2/3 < h < 0 -20h <= 10 and -2/3 < h < 0 h >= -1/2 and -2/3 < h < 0 -1/2 <= h < 0 or (c) -5h >= 5(3h+2) and h > 0 -5h >= 15h+10 and h > 0 -20h >= 10 and h > 0 h <= -1/2 and h > 0 (no solutions here...) And so our solution set, from parts (a) and (b), is once again (-infinity,-2/3) or [-1/2,0). I hope all this helps and that the several different methods do not confuse you. I find I am much more comfortable and confident with a problem where I know I have several different methods I can try to use in case I have trouble using one of the methods; or if an answer I get by one method doesn't seem right and I want to check it by using another method. Write back if you have any further questions on this topic. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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